CBSE Notes, Lectures

CBSE - Physics - Motion

Motion

NCERT In Text

A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

Initial speed of the train, u= 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = - 0.5 m s-2
According to third equation of motion:
v2= u2+ 2 as
(0)2= (25)2+ 2 ( -0.5) s

Where, s is the distance covered by the train



The train will cover a distance of 625 m before it comes to rest.

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